Evaluate $\int\sin^4x\,dx\,$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac14(1-2\cos2x+\cos^22x)+C\\$ (Choice B) B $\dfrac32x-\sin2x+\dfrac18\sin4x+C\\$ (Choice C) C $\dfrac14\Big(\dfrac12x-\sin2x+\dfrac14\sin4x\Big)+C\\$ (Choice D) D $\dfrac14\Big(\dfrac32x-\sin2x+\dfrac18\sin4x\Big)+C\\$
Solution: This does not have an odd power of either $~\sin x~$ or $~\cos x\,$, so we rewrite the even power of $~\sin x~$ using the double-angle identity for $~\cos 2x\,$. $ \cos2x=1-2\sin^2x~~~~~\Rightarrow~~~~~\sin^2x=\dfrac{1-\cos2x}2$ $\begin{aligned} \int(\sin^2x)^2\, dx&= \int\Bigg(\dfrac{1-\cos2x}{2}\Bigg)^2\,dx \\\\ &=\dfrac14\int\big(1-2\cos2x+\cos^22x\big)\,dx \end{aligned}$ Now we have an even power of $~\cos \,$. We use the identity $~~~\cos^22x=\dfrac12\big(1+\cos4x\big)~~~-$ another version of the double-angle identity for $~\cos-~~~$ to obtain all odd powers. $\begin{aligned} \int(\sin^2x)^2\, dx&=\dfrac14\int\big(1-2\cos2x+\cos^22x\big)\,dx \\\\ &=\dfrac14\int\bigg(1-2\cos2x+\dfrac12\Big(1+\cos4x\Big)\bigg)\,dx \\\\ &=\dfrac14\int\Big(\dfrac32-2\cos2x+\dfrac12\cos4x\Big)\,dx \\\\ &=\dfrac14\Big(\dfrac32x-\sin2x+\dfrac18\sin4x\Big)+C \end{aligned}$